Question

If $a=2+\sqrt{3}$ and $b=2-\sqrt{3}$ then $\frac{1}{a^2}-\frac{1}{b^2}$

• A. $14$
• B. $-12$
• C. $-8\sqrt{3}$
• D. $\sqrt{3}$

Answer 1

The correct option is C $-8\sqrt{3}$

$a=2+\sqrt{3}$

$\frac{1}{a}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

$\frac{1}{a^2}={\left(2-\sqrt{3}\right)}^2=-4+3-4\sqrt{3}$

$b=2-\sqrt{3}$

$\frac{1}{b}=\frac{1}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}$

$=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}$

$\frac{1}{b^2}={\left(2+\sqrt{3}\right)}^2=4+3+4\sqrt{3}$

$\frac{1}{a^2}-\frac{1}{b^2}=4+3-4\sqrt{3}-\left(4+3+4\sqrt{3}\right)$

$=-4\sqrt{3}--4\sqrt{3}$

$=-8\sqrt{3}$